Question

$\displaystyle\lim_{x\to \frac{\pi}{2}} \frac{\left[1- \tan\left(\frac{x}{2}\right)\right] \left[1-\sin x\right]}{\left[1+\tan\left(\frac{x}{2}\right)\right]\left[\pi-2x\right]^{3}} $ is

• A. $\infty$
• B. $\frac{1}{8}$
• C. 0
• D. $\frac{1}{32}$

Answer 1

Answer: (D)

$\displaystyle \lim_{x \to\frac{\pi}{2}} \frac{\tan\left(\frac{\pi}{4} - \frac{x}{2}\right) .\left(1-\sin x\right)}{\left(\pi-2x\right)^{3}} $
Let $x= \frac{\pi}{2} + y; y \to 0 $
$=\displaystyle \lim_{y\to0} \frac{\tan\left(-\frac{y}{2}\right) .\left(1-\cos y\right)}{\left(-2y\right)^{3}} = \displaystyle \lim_{y\to0} \frac{-\tan \frac{y}{2} 2 \sin^{2} \frac{y}{2}}{\left(-8\right). \frac{y^{3}}{8} .8} $
$ =\displaystyle \lim_{y \to 0} \frac{1}{32} \frac{\tan \frac{y}{2}}{\left(\frac{y}{2}\right)}. \left[\frac{\sin y/2}{y/2}\right]^{2} = \frac{1}{32} $